Question

Calc

Answer 1

\[\sec ^{-1}(e^x)=f(x)\]
\[f'(x) = ?\]
solving for f(x)
Teacher gave these formulas for inverse trig derivatives

for arc sec

\[d/dx(\sec ^{-1}x )=1/|x|\sqrt{(x^2)-1}\]

Answer 2

\[1/(|x|\sqrt{(x^2-1)})\]

Answer 3

I plugged in e^x for everywhere that has an x, idk y i got it wrong

Answer 4

@imqwerty @jimthompson5910

Answer 5

\(\sec^{-1}\left(e^{x}\right) = f(x)\)

\(f(x) = \sec^{-1}\left(e^{x}\right)\)

\(\frac{d}{dx}\left[f(x)\right] = \frac{d}{dx}\left[\sec^{-1}\left(e^{x}\right)\right]\)

\(f'(x) = \frac{1}{|e^x|\sqrt{(e^x)^2-1}}*\frac{d}{dx}\left[e^x\right] \ \text{... Chain Rule}\)

I'll let you finish up

Answer 6

oof, i get it now, the derivative of e^x (derivative of the inside part) is e^x
We have an e^x in the denominator as well, no need for absolute value because e^x will always be positive so they cancel
\[1/((e^{2x}-1)\]

Answer 7

oops that was supposed to be under square root

Answer 8

the e^2x - 1 part

Answer 9

yes you are correct

Answer 10

thx jimthompson : )

Answer 11

no problem